Let's solve the problem using the concept of joint variation.

Given:

• Horsepower $hp$ varies jointly with the speed $N$ and the cube of the diameter $d$.
• Mathematically, this can be expressed as:
  $hp = k \cdot N \cdot d^3$ where $k$ is the constant of proportionality.

Step 1: Find the constant $k$.

From the information given:

• $hp = 36$ horsepower
• $N = 75$ rpm
• $d = 2$ inches

Substitute these values into the equation: $36 = k \cdot 75 \cdot (2)^3$ $36 = k \cdot 75 \cdot 8$ $36 = 600k$ $k = \frac{36}{600} = 0.06$

Step 2: Use $k$ to find the required diameter $d$ for the new conditions.

Now, we need to find $d$ when:

• $hp = 38$ horsepower
• $N = 80$ rpm

Substituting these values into the equation with the constant $k$ already found: $38 = 0.06 \cdot 80 \cdot d^3$ $38 = 4.8 \cdot d^3$ $d^3 = \frac{38}{4.8} \approx 7.917$ $d \approx \sqrt[3]{7.917} \approx 1.996 \text{ inches}$

Final Answer:

The diameter of the shaft must be approximately 1.996 inches.

Would you like more details or have any questions?

Here are some related questions:

1. How is the constant of proportionality $k$ determined in a joint variation problem?
2. What are some examples of joint variation in real-world applications?
3. How would the result change if the speed was doubled while keeping the horsepower constant?
4. What is the relationship between the diameter and horsepower in this context?
5. What happens to the required diameter if both speed and horsepower are doubled?

Tip: In joint variation problems, always solve for the constant of proportionality first to simplify calculations for new conditions.